1The nth particle enters just the container because the weight of the particle is more than the total electrostatic repulsive force ...
I'll try to get a simplifed mathematical expr. after sometime ...
electrically charged drops of mercury fall from altitude h into a spherical metal vessel of radius R. in the upper part of which there is a small opening. The mass of each drop is m & charge is Q.What is the number 'n' of last drop that can enter the sphere.Given that (n+1)th
drop just fails to enter the sphere?
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7 Answers
49balance forces! will try to come up with a solution in few hours!!!
btw akhil..u in 11 or 12???
149let the electrc field at the hole due to n drops be E
thus balancing force on the (n+1)th drop we get,
QE=mg
rest of the problem is very easy .. only we need not to get confused!!
total charge in n drops = nQ
so since the shell is metallic, and the hole is small such that no electric lines of force escapes [i am trying to cheat in order to work less!! ;)]
so charge induced in inner surface is (-nQ)
so charge appearing on outer surface is nQ
field on the hole is nQ8πεoR2 [not nQ4πεoR2...why??]
thus equating we get...
mg=nQ28πεoR2
that gives,
n=8πεoR2mgQ2
49why is the field at the hole nQ8πεoR2 and not nQ4πεoR2 ???
lets see who can answer this? ;)
49now lets try out some googlies here!!
WHAT IF THE SPHERE IS NON-CONDUCTING??
take each drop is spherical and has a radius of r ... !!
hint: i can think about nothing else other than using solid angle concept!
66@subho
balancing forces won't give you the desired result. Can you figure out why?
1Anant sir .... just a doubt ..... the electric field intensity due to one falling charge on another inside the container is not const . but changing .... as the distance between the 2 is continuosly decreasing ... [ charges r not static ]
So how can we apply the normal algebraic eqn. ????
